题目内容
| lim |
| n→∞ |
| 1 |
| n2+1 |
| 2 |
| n2+1 |
| 3 |
| n2+1 |
| 2n |
| n2+1 |
分析：首先求极限
(
+
+
+…+
)发现式子上面各项是等差数列，即可求和，得到
容易求得它的极限为2，即为答案．
| lim |
| n→∞ |
| 1 |
| n2+1 |
| 2 |
| n2+1 |
| 3 |
| n2+1 |
| 2n |
| n2+1 |
| 2n2+n |
| n2+1 |
解答：解：设A=
+
+
+…+
=
=
所以
(
+
+
+…+
)=
A=
=2
故答案为2．
| 1 |
| n2+1 |
| 2 |
| n2+1 |
| 3 |
| n2+1 |
| 2n |
| n2+1 |
| 1+2+3+…+2n |
| n2+1 |
| 2n2+n |
| n2+1 |
所以
| lim |
| n→∞ |
| 1 |
| n2+1 |
| 2 |
| n2+1 |
| 3 |
| n2+1 |
| 2n |
| n2+1 |
| lim |
| n→∞ |
| lim |
| n→∞ |
| 2n2+n |
| n2+1 |
故答案为2．
点评：此题主要考查极限及其运算的问题，其中涉及到等差数列的求和问题，属于综合题，有一定的计算量，为中档题．
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